I have a random variable $X$ taking values in the interval $[0,1]$ with mean 0.2 and median 0.3. What are the lower and upper bounds of the set of possible variances of $X$?
I am able to solve this for a distribution having equal mass at two points (it gives the highest variance for points $\{0,0.4\}$ and the lower bound is the variance of the distribution for points $\{0.1,0.3\}$). Is it possible to somehow reduce the initial problem to this case?
I would be grateful for your help.
Best Answer
Sorry to come 6 years late to the party, but perhaps this will be useful for someone looking at a similar problem. Here is one naive approach to derive at least some set of bounds.
Let's take a discrete uniform distribution on the set $S = \{a, 0.3, b\}$ with $a \le 0.3 \le b$, which ensures that any random variable $X$ drawn from this distribution will have median $0.3$ as requested. We must have mean of $0.2$, which implies $$ 0.2 = \mathbb{E}[X] = \frac{a+b+0.3}{3} \iff a+b = 0.3 $$ Now we substitute $a = 0.3 - b$, and our set basically becomes $S = \{0.3 - b, 0.3, b\}$, with $b \ge 0.3$ and $a = 0.3 - b \le 0.3$.
Since $X$ must take values in $[0,1]$, we have $b \le 1$ and $a = 0.3 -b \ge 0$, which effectively forces $b = 0.3$ and has $S = \{0, 0.3, 0.3\}$ with variance $0.06$.
Using other distributions may result in different values. Let's make another attempt by considering a distribution on two points with different probabilities. Then, to make the median $0.3$, we must have one point at any $x \in [0,1]$ with probability $p< 1/2$, and another one at at $0.3$ with probability $1-p>1/2$. Then, $$ \mathbb{E}[X] = xp + 0.3(1-p) = 0.2 \iff 0.3 p - xp = 0.1 $$ and the variance would be $$ v(x,p) = \mathbb{Var} X = (1-p)(0.3-0.2)^2 + p(x-0.2)^2 = p(x-0.2)^2 + 0.01(1-p). $$
So now one has to find the extrema of $v(x,p)$ subject to $0.3p - xp = 0.1$. The easiest way to do this is to use the constraint and solve for $p$, getting $$ p = \frac{0.1}{0.3-x} $$
Since we must have $p \ge 0$, this forces $x < 0.3$, and since we must have $p \le 1/2$, this implies $x \le 0.1$, so we have $x \in [0, 0.1]$.
Plugging it into the variance, getting $$ v(x) = \frac{0.1(x-0.2)^2 - 0.001}{0.3-x} + 0.01 = 0.02(1-5x) $$ which needs to be minimized subject to $x \in [0,0.1]$. So $$ \begin{split} x_\min = 0.1 &\implies p = \frac12 &\implies v_\min = 0.01 \\ x_\max = 0 &\implies p = \frac13 &\implies v_\max = 0.02 \\ \end{split} $$
So overall we now have $v_\min \le 0.01$ and $v_\max \ge 0.06$.